Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Forum Expert

Joined:	28 Feb 2007
Post:	2173

27 Jul 2008 14:39:41 IST

0 People liked this

930

Easy NT - School level

None

1. Find all n such that 2⁸+2¹¹+2ⁿ is a perfect square

2. If b is an even number such that $\frac{b^n-1}{b-1}$ is a perfect square, then prove that b must be divisible by 8

Share this article on:

Comments (13)

Sahil Gupta

Blazing goIITian

Joined: 10 Jun 2007

Posts: 985

27 Jul 2008 15:00:19 IST

Like 0 people liked this

wait. trying hard

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

27 Jul 2008 16:46:26 IST

Like 0 people liked this

Re:Easy NT - School level

Sahil Gupta

Blazing goIITian

Joined: 10 Jun 2007

Posts: 985

27 Jul 2008 17:11:03 IST

Like 0 people liked this

@rahul i don't think we can use hit and trial in such questions. and moreover this gave you just one value of n.

but well tried for you are a 10^th grader

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

27 Jul 2008 17:13:36 IST

Like 2 people liked this

solving 2, leaving 1 for rahul .D:

it's basically an easy problem !!

first write

$\frac{b^n-1}{b-1}=1+b+b^2 +...+b^{n-1}$ ....(1)

Now b being an even no , RHS is odd , which is given to be perfect square . So , without any loss of generality, we may write (1)as = (2c+1)^2=4c^2+4c+1 for some integer 'c' .

so we get ultimately $b(1+b+b^2+...+b^{n-2})=4c(1+c)$ ..(2)

Now consider two cases :

Case 1: c is even =2m (say)

in that case , b being even , the bracketed portion is odd , so it can't be divisible by 8( RHS =8 * odd factor), which means b is divisible by 8.

Case 2: c is odd

In that case also ( 1+c ) = even = 2n say , so in the RHS we get 8 * odd factor , so b being even , it must be divisible by 8.

(Q.E.D.)

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

27 Jul 2008 17:43:32 IST

Like 0 people liked this

@feynmann: I wish you had left even this easy one for the school boys to solve. Arent there bigger fish to catch in this forum?

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

27 Jul 2008 17:45:08 IST

Like 0 people liked this

and anybody for more rigorous soln for the 1st one?

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

27 Jul 2008 19:34:10 IST

Like 0 people liked this

Re:Easy NT - School level

Rajat Sen

Blazing goIITian

Joined: 7 Aug 2007

Posts: 533

28 Jul 2008 14:51:46 IST

Like 0 people liked this

2^8+2^11+2^n = k^2

so, 2^8(1+2^3 + 2^(n-8)) = k^2

so, 1+2^3 + 2^(n-8) = q^2 for some q (-- I

so, 2^(n-8) = (q+3)(q-3)

so, for some r and s 2^r - 2^s = 6 [ r+s = n-8]

so, 2^s(2^r-s - 1) = 6 [ 6 = 2*3]

so, s =1

r = 3

n - 8 = 4

so, n= 12

please rate

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

28 Jul 2008 15:52:28 IST

Like 0 people liked this

How do you guys know that 2^n-8 is an integer?

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

28 Jul 2008 19:27:22 IST

Like 0 people liked this

Re:Easy NT - School level

Rajat Sen

Blazing goIITian

Joined: 7 Aug 2007

Posts: 533

29 Jul 2008 12:23:49 IST

Like 0 people liked this

suppose n<8

2^8 +2^11 + 2^n = k^2

2^8(2^3 +1 + 2^(n-8)) = k^2

so , 8 + 1+ 2^(n-8) = (k/16)^2

so, 9 = (k/16)^2 - 2^(n-8)

let n - 8 = -m ( m being a positive integer)

so, 9 = (k/16)^2 - 2^(-m) ( multiplying by 2^(m))

1/ 2^m = (k/16 - 3)(k/16 +3)

let m = r+s

so, 1/2^r - 1/2^s = 6

so, (2^(s-r) - 1)/2^s = 6

the only odd number in rhs = 3

so, 2^(s-r) - 1 = 3

but that implies 2^(-s) = 2

so, s = -1

but s - r = 2

so, r = -3

but m =(r+s) > 0

so, this is a contradiction.

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

30 Jul 2008 15:52:14 IST

Like 1 people liked this

We have $2^8+2^{11}+2^n = 9.2^8+2^n = (2.2^4)^2+2^n$

For a perfect square a², the least difference to a higher square is 2a+1

Hence $2^n \ge 2(2^4.3) + 1 = 97$

That means n>=7. n=7 is ruled out as the highest power of 2 that divides LHS would be odd and hence cannot be perfect square.

So, n>=8 and the expression is now 2⁸(9+2^n-8) = 2⁸(9+2^p)

So, we must have 9+2^p is a perfect square

Hence 9+2^p = m² and so (m-3)(m+3) = 2^p

So, m-3 and m+3 are both powers of 2. Let m-3 = 2^u and m+3 = 2^v such that u+v = p

Then 6 = 2^u(2^v-u-1) gives u =1, v = 3 and hence p = 4 and n = 12 as the only solution

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

30 Jul 2008 15:58:02 IST

Like 1 people liked this

Also, the question $\frac{q^n-1}{q-1}$ being a perfect square when q is even only if q is a multiple of 8:

Every odd square that is not a multiple of 8 is of the form 8k+1

Hence 1+b+b²+b³+...+bⁿ is of the form 8k+1 or

b+b²+b³+...+bⁿ = b(1+b+b²+...+b^n-1) is divisible by 8 and since the bracketed expression is an odd number, b is divisible by 8

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team